Question : If $\sin A=\frac{1}{2}$, then the value of $(\tan A+\cos A)$ is:
Option 1: $\frac{2}{3 \sqrt{3}}$
Option 2: $\frac{3}{2 \sqrt{3}}$
Option 3: $\frac{5}{2 \sqrt{3}}$
Option 4: $\frac{5}{3 \sqrt{3}}$
Correct Answer: $\frac{5}{2 \sqrt{3}}$
Solution : Given: $\sin A=\frac{1}{2}$ ⇒ $\sin A=\sin 30^\circ$ ⇒ $ A= 30^\circ$ $\therefore (\tan A+\cos A)$ = $(\tan 30^\circ+\cos 30^\circ)$ = $\frac{1}{\sqrt3}+\frac{\sqrt3}{2}$ = $\frac{2+3}{2\sqrt3}$ = $\frac{5}{2\sqrt3}$ Hence, the correct answer is $\frac{5}{2\sqrt3}$.
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Question : Find the value of $\sqrt{\frac{1-\tan A}{1+\tan A}}$.
Option 1: $\sqrt{\frac{1+\sin 2 A}{\cos 2 A}}$
Option 2: $\sqrt{\frac{1-\sin 2 A}{\cos 2 A}}$
Option 3: $\sqrt{\frac{1+\sin A}{\cos A}}$
Option 4: $\sqrt{\frac{1-\sin A}{\cos A}}$
Question : If $\sin \theta+\cos \theta=\frac{\sqrt{11}}{3}$, then the value of $(\cos \theta-\sin \theta)$ is:
Option 1: $\frac{\sqrt{5}}{3}$
Option 2: $\frac{7}{3}$
Option 3: $\frac{5}{3}$
Option 4: $\frac{\sqrt{7}}{3}$
Question : If $\sin A=\frac{2}{3}$, then find the value of (7 – tan A)(3 + cos A).
Option 1: $\frac{61}{3}+\frac{17}{3 \sqrt{5}}$
Option 2: $\frac{61}{3 \sqrt{5}}+\frac{17}{3}$
Option 3: $\frac{61}{3}+\frac{17}{\sqrt{5}}$
Option 4: $\frac{61}{3}-\frac{17}{3 \sqrt{5}}$
Question : If $\sin (A-B)=\sin A \cos B–\cos A\sin B$, then $\sin 15°$ will be:
Option 1: $\frac{\sqrt{3}+1}{2\sqrt{2}}$
Option 2: $\frac{\sqrt{3}}{2\sqrt{2}}$
Option 3: $\frac{\sqrt{3}–1}{–\sqrt{2}}$
Option 4: $\frac{\sqrt{3}–1}{2\sqrt{2}}$
Question : If $\sin \theta+\cos \theta=\frac{\sqrt{11}}{3}$, then what is $\sin \theta-\cos \theta$?
Option 1: $\frac{\sqrt{7}}{4}$
Option 2: $\frac{\sqrt{7}}{3}$
Option 3: $\frac{\sqrt{5}}{3}$
Option 4: $\frac{\sqrt{5}}{2}$
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